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Let $\Delta PQR$ be an acute-angled triangle. $D$ and $E$ are points on $PQ$ such that $PD:DE:EQ = 1:1:1$. $H$ is on $RP$ such that $RH:HP=2:1$. $G$ is on $RQ$ such that $RG:GQ=2:1$. $RD$ meets $HQ$ at point $I$ while $PG$ meets $RE$ at point $J$. Prove that $IJ$ is parallel to $PQ$.

I have sketched the graph already. Diagram in here

It seems that it can be done easily if we use vector or coordinates. However, is there a way to prove it by using similar triangles and/or other geometry properties only? I have tried to add $HG$ and it can be proven that $HG$ is parallel to $PQ$, but I cannot move on from here. If possible, can you give me some hints to approach this question? I don't want to get the solutions directly, guidance is much appreciated. Thank you!

Andrei
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Am_2099
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  • @MathLover : Thank you for your reply. I seldom used Menelaus' Theorem so I did not try it before. Let me attempt the question again! Meanwhile, would you mind answering the question instead of leaving comments? If there are no other replies, I can then select yours! – Am_2099 Feb 15 '22 at 15:36
  • ok I will. Let me know if you get stuck. – Math Lover Feb 15 '22 at 15:38

1 Answers1

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Apply Menelaus' theorem in $\triangle PRD$ with secant $HQ$ and show $ \displaystyle \frac{DI}{IR} = \frac 13$. Similarly show $ \displaystyle \frac{EJ}{JR} = \frac 13$.

Using $ ~\displaystyle \frac{DI}{IR} = \frac{EJ}{JR}$, conclude $~IJ \parallel PQ$.

Math Lover
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